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loris
2 months ago
9

At Monroe High School, 62% of all students participate in after school sports and 11% participate in both after school sports an

d student council. What is the probability that a student participates in student council given that the student participates in after school sports?
Mathematics
2 answers:
zzz [12.3K]2 months ago
4 0
Hello!

Before you tackle any problems, it's essential to designate which scenario represents event A and which corresponds to event B. I usually follow the order they are presented in the question, so:

A = S<span>tudent participates in student council
B = S</span><span>tudent participates in after school sports

Any problem that mentions "given" in the question will need to refer to </span>P(A | B)<span> = P(</span>A ∩ B)/P(B). P(A | B) essentially represents the "probability of event A, given that event B has occurred." Meanwhile, P(A ∩ B) denotes the likelihood of both A and B taking place, while P(B) signifies just the probability of event B happening. All required information has been provided, so:

P(A | B) = P(A ∩ B)/P(B)
P(A | B) = 11% / 62%
P(A | B) = 0.11 / 0.62
P(A | B) = 0.18

Thus, there is approximately an 18% probability that <span>a student is involved in student council, given participation in after school sports.

I hope this was helpful!:-)</span>
PIT_PIT [12.4K]2 months ago
4 0

Answer with Step-by-step explanation:

At Monroe High School, 62% of students engage in after school sports while 11% are involved in both after school sports and student council.

A: students attending after school sports

P(A)=0.62

B: students in student council

A∩B: students participating in both after school sports and student council.

P(A∩B)=0.11

B/A: student involvement in student council, given that they are part of after school sports

Bayes' theorem states that

P(A∩B)=P(B/A)×P(A)

0.11=P(B/A)×0.62

⇒ P(B/A)=0.18

Therefore, the probability that a student participates in student council, knowing they engage in after school sports, is:

0.18

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